The Line $y=-\frac{1}{5} X$ Is A Normal To The Curve $y=x \sqrt{x}-x+c$. Find The Value Of $ C C C [/tex].

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Introduction

In mathematics, the concept of a normal line to a curve is crucial in understanding the properties and behavior of functions. A normal line is a line that is perpendicular to the tangent line at a given point on the curve. In this article, we will explore how to find the value of c in the equation of a curve, given that a specific line is a normal to the curve.

The Curve and the Normal Line

The curve is given by the equation $y=x \sqrt{x}-x+c$, and the normal line is represented by the equation $y=-\frac{1}{5} x$. To find the value of c, we need to use the fact that the normal line is perpendicular to the tangent line at a given point on the curve.

Finding the Slope of the Tangent Line

To find the slope of the tangent line, we need to find the derivative of the curve with respect to x. Using the product rule and the chain rule, we can find the derivative of the curve as follows:

dydx=ddx(xx−x+c)\frac{dy}{dx} = \frac{d}{dx} (x \sqrt{x}-x+c)

dydx=x+x12x−1\frac{dy}{dx} = \sqrt{x} + x \frac{1}{2\sqrt{x}} - 1

dydx=32x−1\frac{dy}{dx} = \frac{3}{2} \sqrt{x} - 1

The slope of the tangent line is given by the derivative of the curve, which is $\frac{3}{2} \sqrt{x} - 1$.

Finding the Slope of the Normal Line

The slope of the normal line is given by the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is:

mnormal=−132x−1m_{normal} = -\frac{1}{\frac{3}{2} \sqrt{x} - 1}

Equating the Slopes

Since the normal line is perpendicular to the tangent line, their slopes are negative reciprocals of each other. Therefore, we can equate the slopes of the two lines as follows:

−15=−132x−1-\frac{1}{5} = -\frac{1}{\frac{3}{2} \sqrt{x} - 1}

Solving for x

To solve for x, we can start by cross-multiplying the equation:

−15(32x−1)=−1-\frac{1}{5} (\frac{3}{2} \sqrt{x} - 1) = -1

−310x+15=−1-\frac{3}{10} \sqrt{x} + \frac{1}{5} = -1

−310x=−65-\frac{3}{10} \sqrt{x} = -\frac{6}{5}

x=4\sqrt{x} = 4

x=16x = 16

Finding the Value of c

Now that we have found the value of x, we can substitute it into the equation of the curve to find the value of c:

y=xx−x+cy = x \sqrt{x} - x + c

y=1616−16+cy = 16 \sqrt{16} - 16 + c

y=64−16+cy = 64 - 16 + c

y=48+cy = 48 + c

Since the point (16, y) lies on the curve, we can substitute x = 16 and y = -\frac{1}{5} x = -\frac{16}{5} into the equation of the curve to find the value of c:

−165=48+c-\frac{16}{5} = 48 + c

c=−165−48c = -\frac{16}{5} - 48

c=−165−2405c = -\frac{16}{5} - \frac{240}{5}

c=−2565c = -\frac{256}{5}

Therefore, the value of c is -\frac{256}{5}.

Conclusion

Q: What is a normal line to a curve?

A: A normal line to a curve is a line that is perpendicular to the tangent line at a given point on the curve.

Q: How do you find the slope of the tangent line?

A: To find the slope of the tangent line, you need to find the derivative of the curve with respect to x. This involves using the product rule and the chain rule to differentiate the curve.

Q: What is the relationship between the slope of the tangent line and the slope of the normal line?

A: The slope of the normal line is the negative reciprocal of the slope of the tangent line.

Q: How do you find the value of c in the equation of a curve, given that a specific line is a normal to the curve?

A: To find the value of c, you need to use the fact that the normal line is perpendicular to the tangent line at a given point on the curve. You can substitute the value of x into the equation of the curve and solve for c.

Q: What is the value of c in the equation of the curve y = x \sqrt{x} - x + c, given that the line y = -\frac{1}{5} x is a normal to the curve?

A: The value of c is -\frac{256}{5}.

Q: How do you know that the line y = -\frac{1}{5} x is a normal to the curve y = x \sqrt{x} - x + c?

A: You can use the fact that the normal line is perpendicular to the tangent line at a given point on the curve. By finding the slope of the tangent line and the slope of the normal line, you can determine if they are perpendicular.

Q: What is the significance of finding the value of c in the equation of a curve?

A: Finding the value of c in the equation of a curve can help you understand the behavior and properties of the curve. It can also help you to identify the curve and its characteristics.

Q: How do you apply the concept of a normal line to a curve in real-world problems?

A: The concept of a normal line to a curve can be applied in various real-world problems, such as:

  • Finding the maximum or minimum value of a function
  • Determining the rate of change of a quantity
  • Analyzing the behavior of a system or a process

Q: What are some common applications of the concept of a normal line to a curve?

A: Some common applications of the concept of a normal line to a curve include:

  • Physics: to analyze the motion of objects and determine the forces acting on them
  • Engineering: to design and optimize systems and processes
  • Economics: to analyze the behavior of economic systems and determine the effects of policy changes

Q: How do you extend the concept of a normal line to a curve to higher dimensions?

A: To extend the concept of a normal line to a curve to higher dimensions, you need to use the concept of a normal plane to a surface. This involves finding the normal vector to the surface at a given point and using it to determine the equation of the normal plane.

Q: What are some challenges and limitations of the concept of a normal line to a curve?

A: Some challenges and limitations of the concept of a normal line to a curve include:

  • Finding the derivative of the curve, which can be difficult or impossible in some cases
  • Determining the slope of the tangent line, which can be difficult or impossible in some cases
  • Applying the concept to higher dimensions, which can be complex and challenging.